How to explain the Maxwell Boltzmann distribution graph (physically)?

Semoi's answer is good. But since you say

I would prefer an intuitive explanation rather than a mathematical one

to express it more simply, and without formulae, the Maxwell distribution is the chi distribution with three degrees of freedom (the components of velocity in Euclidean space). The chi distribution is the distribution of the positive square root of the sum of squares of a set of independent random variables each following a standard normal distribution. The normal distribution is the result of the central limit theorem, which basically says that when you have loads of identical random variables added together, the result tends to a normal distribution.

So, Maxwell assumed that the underlying distribution of velocities should be random, and it should result from very large numbers of collisions, and consequently it should be normally distributed in each direction. The Maxwell distribution graph follows from that, using standard calculations.

Note on second part: I do not understand why you ask why some collisions are more favoured. All collisions just have the effect of randomising the velocity distribution both in magnitude and direction. They cannot change overall energy, which determines the mean of the normal distribution in each direction.

The phase space argument put forward by Shaswata is correct. From a thermodynamical perspective this is is merely the result of the of the three dimensional ideal gas relation $$ P_{v} = \frac{n_{v}}{N_{total}} = \frac{1}{Z} \exp\left({-\frac{E_{v}}{k_B T}}\right) $$ where $E_{v}= \frac{m}{2}v^2$ is the energy, $n_{v}$ is the number of particles with velocity $v=|\vec v|$, $N_{total} = \sum_v n_{v}$ is the total number of particles, and $Z_ = \sum_v \exp\left({-\frac{E_{v}}{k_B T}}\right)$ is the partition function. You will find proper derivatives in many thermodynamics books. However, for me this is not an intuitive or conceptional argument. Thus, here is how I like to obtain the Maxwell-Boltzmann distribution.

1. Each component of the velocity vector, $\{v_x, v_y, v_z\}$, is normally distributed: This follows from the fact that we obtain a normal distribution (= Gaussian), if we have "many" independent random processes which contribute "equally" to the observable variable. For justification check out the Poisson theorem and how the Poissonian approached the normal distribution. Or check out the so called Central limit theorem.
2. The square of the velocity, $v^2 = \sum_{i=x,y,z} v_i^2$, is distributed as $\chi^2_3$ (omitting the normalisation constant): This simply is the definition of the $\chi^2$ distribution. Most often the definition reads: If $Z_i \sim N(0, 1)$ then $Y=\sum_{i=1}^k Z_i^2 \sim \chi^2_{k}$. Note that the tilde "$\sim$" should be read as "is distributed as $\ldots$" and that the probability density function is given by $f_Y \propto \sqrt{Y} \exp\left(-Y/2\right)$, for $k=3$.
3. However, we are not interested in distribution of $Y = \vec v^2/\sigma_v^2$, but we'd like to know the distribution of $v=|\vec v| = \sigma_v \sqrt{Y}$. Hence, we have to transform the $\chi_3^2$ distribution. However, this is straight forward. Using the transformation law (for continuous random variables) \begin{align} f_v \; dv &= f_y \cdot \left|\frac{dY}{dv}\right| \;dv \\ &= f_y \cdot \frac{2 v}{\sigma_v^2} \; dv \propto v^2\exp\left( -\frac{v^2}{2\sigma_v^2}\right) \; dv \end{align} where $f_v$ is the so called probability density function for the random variable $v$, we obtain the Maxwell Boltzmann distribution.

In conclusion, the Maxwell Boltzmann distribution of $|\vec{v}|$ is a direct consequence of the fact that each velocity component, e.g. $v_x$, is normally distributed.

All the collisions are equally likely. The maxwell distribution can be derived from the Boltzmann probabilities- namely the probability of a molecule having energy $E$ is $e^{-\beta E}$. Once you can show this it is not so hard to derive the maxwell relation since for any velocity $v$ the number of states is proportional to $e^{-\beta \frac{1}{2}mv^2}$. It is also proportional to $4\pi v^2$ since in the 3D space $v_x,v_y,v_z$ can take any value given that $v_x^2+v_y^2+v_z^2=v^2$ which is basically proportional to the number of points on the surface of a sphere of radius $v$.

So the real question is how the particles achieve through constant molecular collisions the Boltzmann distribution. Why the probability of a molecule having energy $\frac{1}{2}mv^2$ is $e^{-\beta \frac{1}{2}mv^2}$.

Say in equilibrium the number of particles having velocity $v_i$ is $n_i$. We are restricted to a system where the $\sum n_i=N$ and the total energy $\sum n_i \frac{1}{2}mv_i^2=E$. The state having the maximum multiciplity is,

$$ \mathrm{arg\,max}_{n_1,n_2,\dots n_k}\frac{N!}{n_1!n_2!\dots n_k!} \qquad \sum n_i=N,\sum n_i \frac{1}{2}mv_i^2=E$$

We know that for the maximum multiplicity without the energy constraint is when all the $n_i$ are equal i.e. $n_i = N/k \quad\forall i$.

Hence it makes sense that when we have a very large energy ($v_j>v$) for one of the states the corresponding $n_j$ will be small. Otherwise, the other $n_i$ will be too small due to the energy constraint. And when we have one of the $n_i$ to be too large and the others are too small we have a small multiplicity.

Similarly, a low energy will correspond to a high number of molecules in that state.

Hence when you let all kinds of collisions to take place the state that is most likely ends up being the one in which most of the particles have low energy. This can be seen from if you consider the collision of molecules as well. Say the two molecules have final velocities $v1,v_2$ and initial velocities as $u_1,u2$.

From the conservation of energy,

$$v_1^2+v_2^2=u_1^2+u_2^2=V^2$$

Basically, the solution space is the positive quarter of a circle of radius $V$. Now it is not hard to see that for $v_2>V/2$ the number of cases is proportional to $1/3$ (subtending an angle of $\pi/6$). On the other hand for $v_2<V/2$ we have 2 times more cases (subtending an angle of $\pi/3$). In other words $v_1^2$ and $v_2^2$ counterbalance each other. What happens as a result is that $v_1,v_2$ has a greater chance of reducing than of increasing.